Where does this problem come from? Could you provide more context, also to convince us that this is no homework? Also, why are there a $v^Tv$ and a $w^Tw$ in the equations when you know that they are both equal to $1$?
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Federico PoloniOct 16 '12 at 8:26

Hi, Sorry for the typo. It should be $vv^T$ and $ww^T$. The problem may look simple as if it is a homework, but it's not, and I think it's not trivial, at least to me. This is part of my attempt to minimize $\sum_{\sigma}|v^{\dagger}\sigma w|^2$ with Lagrange multiplier. Here {\sigma} are tensor products of some Pauli matrices, and $v$, $w$ are two orthonormal pure state. It is needed to prove another conjecture for my research project in quantum entanglement. I don't even know if it holds although random test suggest it does.
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Minh TranOct 18 '12 at 7:10

1 Answer
1

Your equations say that $Aw$ and $Av$ are both contained in $U=\operatorname{span}(v,w)$, therefore $U$ is an invariant subspace of $A$. You can get all two-dimensional invariant subspaces by taking $U=\operatorname{span}(x_1,x_2)$, where $x_1$ and $x_2$ are eigenvectors of $A$ (proof: consider $A$ restricted to the subspace $U$; it is a symmetric linear operator, so it has two eigenvalues which are also eigenvalues of $A$).

So all solutions must be of the form $v=\alpha x_1 +\beta x_2$ and $w=\gamma x_1+\delta x_2$, where $x_1$ and $x_2$ are two eigenvectors of $A$. Making this ansatz the problem becomes a $2\times 2$ one in $\alpha,\beta,\gamma,\delta$ and should be easy to solve explicitly.

Thank you for your answer. It is not so clear to me why the two eigenvalues of $A$ restricted to $U=span{v,w}$ are also eigenvalues of $A$. Can you explain a bit more?
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Minh TranOct 18 '12 at 10:24

It's the same operator, just seen on a smaller vector subspace. The relation $Au=\lambda u$ still holds, does not depend on the ambient space.
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Federico PoloniOct 18 '12 at 10:40

Thank you for you great idea. I would like to ask you some more. How would you extend your idea to the case where you have two matrices instead of one as above? I have spent me some time to think about it but really I don't see a way.
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Minh TranOct 25 '12 at 3:31

You'd better ask a new question with this second problem. Is the text correct? It is unpleasingly asymmetric.
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Federico PoloniOct 25 '12 at 9:22

Thank you for your advice. I will start a new question. There was indeed some typo anyway.
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Minh TranOct 27 '12 at 8:04